# Time Dilation Problem [closed]

I'm having some trouble using the time dilation formula. Say an astronaut leaves Earth for 10 years, at 0.85c. How much time has passed according to an observer on Earth?

I tried using the following formula:

$$t = \frac{1}{\sqrt{1-(v^2/c^2)}}$$

Any help would be much appreciated! This concept is super confusing to me.

• I think you might be missing a term in your formula. Commented Sep 10, 2014 at 18:40
• Leaves for 10 years in what frame?
– BMS
Commented Sep 10, 2014 at 18:41
• Okay. I wrote the formula exactly how it appeared on my formula sheet, but it's possible I'm missing something. Any idea what term?
– McB
Commented Sep 10, 2014 at 18:42
• Leaves for 10 years relative to the astronaut
– McB
Commented Sep 10, 2014 at 18:44
• Your image equation is a little different from the formula you typed, do you see what is missing? Commented Sep 10, 2014 at 19:33

Your written text says "t = 1/sqrt[1-(v^2/c^2)]". If you used that equation, it's no wonder you got a nonsensical value. In the equation in the scanned image, that's a letter $t$ in the numerator, not the digit $1$.

Also, although it works for this problem, the scanned image should really say something like

$$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}\ .$$

As it's written, it looks like it's expressing a coordinate transformation between the two frames instead of just expressing the time dilation factor, and interpreted as a coordinate transformation the equation would in general be wrong. The general coordinate transformation for the standard configuration is

$$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( t - \frac{vx}{c^2} \right)\ .$$

That equation reduces to the scanned equation if you're only dealing with the world line $x=0$. $x=0$ in this problem expresses that the astronaut is standing still in his coordinate system.